Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

1.5. Random Variables 43

which is the desired result.

Example 1.5.6.LetXhave the discontinuous cdf

FX(x)=

⎧

⎨

⎩

0 x< 0

x/ 20 ≤x< 1

11 ≤x.

Then

P(− 1 <X≤ 1 /2) =FX(1/2)−FX(−1) =

1

4

−0=

1
4
and

P(X=1)=FX(1)−FX(1−)=1−

1

2

=

1

2

.

The value 1/2 equals the value of the step ofFXatx=1.

Since the total probability associated with a random variableXof the discrete
type with pmfpX(x) or of the continuous type with pdffX(x) is 1, then it must be
true that ∑

x∈DpX(x)=1and

∫
DfX(x)dx=1,
whereDis the space ofX. As the next two examples show, we can use this
property to determine the pmf or pdf if we know the pmf or pdf down to a constant
of proportionality.
Example 1.5.7.SupposeXhas the pmf

pX(x)=

{

cx x=1, 2 ,..., 10

0elsewhere,

for an appropriate constantc.Then

1=

∑^10

x=1

pX(x)=

∑^10

x=1

cx=c(1 + 2 +···+ 10) = 55c,

and, hence,c=1/55.

Example 1.5.8.SupposeXhas the pdf

fX(x)=

{

cx^30 <x< 2

0elsewhere,

for a constantc.Then

1=

∫ 2

0

cx^3 dx=c

[

x^4

4

] 2

0

=4c,

and, hence,c=1/4. For illustration of the computation of a probability involving
X,wehave

P

(

1

4

<X< 1

)

=

∫ 1

1 / 4

x^3

4

dx=

255

4096

=0. 06226.